Problem: For a function $g$, we are given that $g(7)=-3$ and $g'(7)=-1$. What's the equation of the tangent line to the graph of $g$ at $x=7$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $y-7=-3(x+1)$ (Choice B) B $y-7=-1(x+3)$ (Choice C) C $y+3=-1(x-7)$ (Choice D) D $y+1=-3(x-7)$
Explanation: The derivative of a function at a point gives the slope of the line tangent to the function's graph at that point. Therefore, $g'(7)$ gives the slope of the tangent line to the graph of $g$ where $x=7$. We are given that $g'(7)=-1$, so the slope of the tangent line is $-1$. Furthermore, we are given that $g(7)=-3$, which means the point of intersection of the tangent line and the graph is $(7,-3)$. To summarize, the tangent line has a slope of $-1$ and it passes through the point $(7,-3)$. We can use the point-slope form of linear equations to find the tangent line equation: $\begin{aligned} y-y_1&=m(x-x_1) \\\\ y-(-3)&=-1(x-7) \\\\ y+3&=-1(x-7) \end{aligned}$ The equation is $y+3=-1(x-7)$.